Timeline for How do I find the algebra representing the projective bundle of a direct sum of line bundles over a projective space?
Current License: CC BY-SA 3.0
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| Sep 4, 2017 at 5:22 | comment | added | 54321user | Also, why can the Hirzebruch surface have such a simpler presentation, as seen here: mathoverflow.net/questions/122952/on-a-hirzebruch-surface ? | |
| Aug 26, 2017 at 22:20 | comment | added | 54321user | I created a bounty for a question related to my comment: math.stackexchange.com/questions/2364401/… | |
| Jul 26, 2017 at 17:00 | comment | added | Sándor Kovács | No. The sheaf of sections of $\mathbb V(\mathscr E)$ is the dual of $\mathscr E$. This is the same phenomenon as with "standard" $\mathrm{Spec}$ and $\mathrm{Proj}$; $\mathrm{Spec}(A)\simeq \mathrm{Spec} B$ if and only if $A\simeq B$, but the analogous statement fails for $\mathrm{Proj}$. The twisting with the line bundle corresponds to a $d$-uple type embedding of the projective scheme/bundle. They are abstractly isomorphic, but have a different embedding. | |
| Jul 26, 2017 at 4:56 | comment | added | 54321user | Does $\mathbb{V}(\mathcal{E})$ remain the same if I twist by a line bundle? | |
| May 6, 2017 at 18:53 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Apr 14, 2017 at 22:35 | vote | accept | 54321user | ||
| Feb 13, 2017 at 20:41 | vote | accept | 54321user | ||
| Feb 13, 2017 at 20:42 | |||||
| Feb 13, 2017 at 3:29 | history | answered | Sándor Kovács | CC BY-SA 3.0 |